david wong

Hey ! I'm David, a security consultant at Cryptography Services, the crypto team of NCC Group . This is my blog about cryptography and security and other related topics that I find interesting.

How to compare password hashes in PHP? May 2015

The wierdness of ==

Do you know what happens when you run this code in PHP?

<?php
var_dump(md5('240610708') == md5('QNKCDZO'));
var_dump(md5('aabg7XSs') == md5('aabC9RqS'));
var_dump(sha1('aaroZmOk') == sha1('aaK1STfY'));
var_dump(sha1('aaO8zKZF') == sha1('aa3OFF9m'));
var_dump('0010e2' == '1e3');
var_dump('0x1234Ab' == '1193131');
var_dump('0xABCdef' == '     0xABCdef');
?>

Check the answer here. That's right, everything is True.

This is because == doesn't check for type, if a string looks like an integer it will first try to convert it to an integer first and then compare it.

More about PHP == operator here

This is weird and you should use === instead.

Even better, you can use hash_equals (coupled with crypt)

Compares two strings using the same time whether they're equal or not.
This function should be used to mitigate timing attacks; for instance, when testing crypt() password hashes.

Here's the example from php.net:

<?php
$expected  = crypt('12345', '$2a$07$usesomesillystringforsalt$');
$correct   = crypt('12345', '$2a$07$usesomesillystringforsalt$');
$incorrect = crypt('apple',  '$2a$07$usesomesillystringforsalt$');

hash_equals($expected, $correct);
?>

Which will return True.

But why?

the hashed strings start with 0e, for example both strings are equals in php:

md5('240610708') = 0e462097431906509019562988736854
md5('QNKCDZO')   = 0e830400451993494058024219903391

because php understands them as both being zero to the power something big. So zero.

Security

Now, if you're comparing unencrypted or unhashed strings and one of them is supposed to be secret, you might have potentialy created the setup for a timing-attack.

Always try to compare hashes instead of the plaintext!

comments (4)

Previous Links April 2015

There is a Link section here that is not very visible, I don't really know how I could show its content on the frontpage here. But here's one way:

May 21th

May 22th

May 23th

May 24th

May 25th

May 27th

May 28th

More

And you can find more on the Links section of this blog

comment on this story

Some research on recovering small RSA private keys April 2015

To make it short, I did some research on the Boneh and Durfee bound, made some code and it worked. (The bound that allows you to find private keys if they are lesser than \(N^{0.292}\))

I noticed that many times, the lattice was imperfect as many vectors were unhelpful. I figured I could try to remove those and preserve a triangular basis, and I went even further, I removed some helpful vectors when they were annoying. The code is pretty straightforward (compare to the boneh and durfee algorithm here)

So what happens is that I make the lattice smaller, so when I feed it to the lattice reduction algorithm LLL it takes less time, and since the complexity of the whole attack is dominated by LLL, the whole attack takes less time.

It was all just theoric until I had to try the code on the plaid ctf challenge. There I used the normal code and solved it in ~3 minutes. Then I wondered, why not try running the same program but with the research branch?

results

That’s right, only 10 seconds. Because I removed some unhelpful vectors, I could use the value m=4 and it worked. The original algorithm needed m=5 and needed a lattice of dimension 27 when I successfully found a lattice of dimension 10 that worked out. I guess the same thing happened to the 59 triplets before that and that’s why the program ran way faster. 3 minutes to 10 seconds, I think we can call that a success!

The original code:

original code

comment on this story

Small RSA private key problem April 2015

/!\ this page uses LaTeX, if you do not see this: \( \LaTeX \)

then refresh the page

Plaid CTF

The third crypto challenge of the Plaid CTF was a bunch of RSA triplet \( N : e : c \) with \( N \) the modulus, \( e \) the public exponent and \( c \) the ciphertext.

plaid rsa triplets

The public exponents \( e \) are all pretty big, which doesn't mean anything in particular. If you look at RSA's implementation you often see \( 3 \), \( 17 \) or other Fermat primes (\( 2^m + 1 \)) because it speeds up calculations. But such small exponents are not forced on you and it's really up to you to decide how big you want your public exponent to be.

But the hint here is that the public exponents are chosen at random. This is not good. When you choose a public exponent you should be careful, it has to be coprime with \( \varphi(N) \) so that it is invertible (that's why it is always odd) and its related private exponent \( d \) shouldn't be too small.

Maybe one of these public keys are associated to a small private key?

I quickly try my code on a small VM but it takes too much time and I give up.

Wiener

A few days after the CTF is over, I check some write-ups and I see that it was indeed a small private key problem. The funny thing is that they all used Wiener to solve the challenge.

Since Wiener's algorithm is pretty old, it only solves for private exponents \( d < N^{0.25} \). I thought I could give my code a second try but this time using a more powerful machine. I use this implementation of Boneh and Durfee, which is pretty much Wiener's method but with Lattices and it works on higher values of \( d \). That means that if the private key was bigger, these folks would not have found the solution. Boneh and Durfee's method allows to find values of private key up to \( d < N^{0.292} \)!

After running the code (on my new work machine) for 188 seconds (~ 3 minutes) I found the solution :)

solution boneh durfee

Here we can see that a solution was found at the triplet #60, and that it took several time to figure out the correct size of lattice (the values of \( m \) and \( t \)) so that if there was a private exponent \( d < N^{0.26} \) a solution could be found.

solution boneh and durfee

The lattice basis is shown as a matrix (the ~ represents an unhelpful vector, to try getting rid of them you can use the research branch), and the solution is displayed.

Boneh and Durfee

Here is the code if you want to try it. What I did is that I started with an hypothesis \( delta = 0.26 \) which tested for every RSA triplets if there was a private key \( d < N^{0.26 } \). It worked, but if it didn't I would have had to re-run the code for \(delta = 0.27\), \(0.28\), etc...

I setup the problem:

# data is our set of RSA triplets
for index, triplet in enumerate(data):

    print "Testing triplet #", index

    N = triplet[0]
    e = triplet[1]

    # Problem put in equation
    P.<x,y> = PolynomialRing(ZZ)
    A = int((N+1)/2)
    pol = 1 + x * (A + y)

I leave the default values and set my hypothesis:

delta = 0.26
X = 2*floor(N^delta)
Y = floor(N^(1/2))

I use strict = true so that if the algorithm will stop if a solution is not sure to be found. Then I increase the values of \( m \) and \( t \) (which increases the size of our lattice) and try again:

solx = -1
m = 2
while solx == -1:
    m += 1
    t = int((1-2*delta) * m)  # optimization from Herrmann and May
    print "* m: ", m, "and t:", t
    solx, soly = boneh_durfee(pol, e, m, t, X, Y)

If no private key lesser than \(N^{delta}\) exists, I try the next triplet. However, if a solution is found, I stop everything and display it.

Remember our initial equation:

\[ e \cdot d = f(x, y) \]

And what we found are \(x\) and \(y\)

if solx != 0:
    d = int(pol(solx, soly) / e)
    print "found the private exponent d!"
    print d

    m = power_mod(triplet[2], d, N)
    hex_string = "%x" % m
    import binascii
    print "the plaintext:", binascii.unhexlify(hex_string)

    break

And that's it!

More?

If you don't really know about lattices, I bet it was hard to follow. But do not fear! I made a video explaining the basics and a survey of Coppersmith and Boneh & Durfee

Also go here and click on the follow button.

comments (3)

Same RSA modulus and correlated public exponents April 2015

Plaid, The biggest CTF Team, was organizing a Capture The Flag contest last week. There were two crypto challenges that I found interesting, here is the write-up of the second one:

You are given a file with a bunch of triplets:

{N : e : c}

and the hint was that they were all encrypting the same message using RSA. You could also easily see that N was the same modulus everytime.

The trick here is to find two public exponent \( e \) which are coprime: \( gcd(e_1, e_2) = 1 \)

This way, with Bézout's identity you can find \( u \) and \( v \) such that: \(u \cdot e_1 + v \cdot e_2 = 1 \)

So, here's a little sage script to find the right public exponents in the triplets:

for index, triplet in enumerate(truc[:-1]):
    for index2, triplet2 in enumerate(truc[index+1:]):
        if gcd(triplet[1], triplet2[1]) == 1:
            a = index
            b = index2
            c = xgcd(triplet[1], triplet2[1])
            break

Now that have found our \( e_1 \) and \( e_2 \) we can do this:

\[ c_1^{u} * c_2^{v} \pmod{N} \]

And hidden underneath this calculus something interesting should happen:

\[ (m^{e_1})^u * (m^{e_2})^u \pmod{N} \]

\[ = m^{u \cdot e_1 + v \cdot e_2} \pmod{N} \]

\[ = m \pmod{N} \]

And since \( m < N \) we have our solution :)

Here's the code in Sage:

m = Mod(power_mod(e_1, u, N) * power_mod(e_2, v, N), N)

And after the crypto part, we still have to deal with the presentation part:

hex_string = "%x" % m
import binascii
binascii.unhexlify(hex_string)

Tadaaa!! And thanks @spdevlin for pointing me in the right direction :)

comment on this story

Plaid CTF April 2015

The Plaid Parliament of Pwning, a security team at Carnegie Mellon University is organizing a CTF right now until tomorrow: http://play.plaidctf.com/

There are two crypto challenges at the moment, and maybe more if someone unlocks one. Have fun!

comments (2)

What are x509 certificates? RFC? ASN.1? DER? April 2015

RFC

So, RFC means Request For Comments and they are a bunch of text files that describe different protocols. If you want to understand how SSL, TLS (the new SSL) and x509 certificates (the certificates used for SSL and TLS) all work, for example you want to code your own OpenSSL, then you will have to read the corresponding RFC for TLS: rfc5280 for x509 certificates and rfc5246 for the last version of TLS (1.2).

rfc ex

x509

x509 is the name for certificates which are defined for:

informal internet electronic mail, IPsec, and WWW applications

There used to be a version 1, and then a version 2. But now we use the version 3. Reading the corresponding RFC you will be able to read such structures:

Certificate  ::=  SEQUENCE  {
    tbsCertificate       TBSCertificate,
    signatureAlgorithm   AlgorithmIdentifier,
    signatureValue       BIT STRING  }

those are ASN.1 structures. This is actually what a certificate should look like, it's a SEQUENCE of objects.

  • The first object contains everything of interest that will be signed, that's why we call it a To Be Signed Certificate
  • The second object contains the type of signature the CA used to sign this certificate (ex: sha256)
  • The last object is not an object, its just some bits that correspond to the signature of the TBSCertificate after it has been encoded with DER

ASN.1

It looks small, but each object has some depth to it.

The TBSCertificate is the biggest one, containing a bunch of information about the client, the CA, the publickey of the client, etc...

TBSCertificate  ::=  SEQUENCE  {
    version         [0]  EXPLICIT Version DEFAULT v1,
    serialNumber         CertificateSerialNumber,
    signature            AlgorithmIdentifier,
    issuer               Name,
    validity             Validity,
    subject              Name,
    subjectPublicKeyInfo SubjectPublicKeyInfo,
    issuerUniqueID  [1]  IMPLICIT UniqueIdentifier OPTIONAL,
                         -- If present, version MUST be v2 or v3
                          subjectUniqueID [2]  IMPLICIT UniqueIdentifier OPTIONAL,
                      -- If present, version MUST be v2 or v3
     extensions      [3]  EXPLICIT Extensions OPTIONAL
                      -- If present, version MUST be v3
}

DER

A certificate is of course not sent like this. We use DER to encode this in a binary format.

Every fieldname is ignored, meaning that if we don't know how the certificate was formed, it will be impossible for us to understand what each value means.

Every value is encoded as a TLV triplet: [TAG, LENGTH, VALUE]

For example you can check the GITHUB certificate here

github cert

On the right is the hexdump of the DER encoded certificate, on the left is its translation in ASN.1 format.

As you can see, without the RFC near by we don't really know what each value corresponds to. For completeness here's the same certificate parsed by openssl x509 command tool:

x509 openssl parsed

How to read the DER encoded certificate

So go back and check the hexdump of the GITHUB certificate, here is the beginning:

30 82 05 E0 30 82 04 C8 A0 03 02 01 02

As we saw in the RFC for x509 certificates, we start with a SEQUENCE.

Certificate  ::=  SEQUENCE  {

Microsoft made a documentation that explains pretty well how each ASN.1 TAG is encoded in DER, here's the page on SEQUENCE

30 82 05 E0

So 30 means SEQUENCE. Since we have a huge sequence (more than 127 bytes) we can't code the length on the one byte that follows:

If it is more than 127 bytes, bit 7 of the Length field is set to 1 and bits 6 through 0 specify the number of additional bytes used to identify the content length.

(in their documentation the least significant bit on the far right is bit zero)

So the following byte 82, converted in binary: 1000 0010, tells us that the length of the SEQUENCE will be written in the following 2 bytes 05 E0 (1504 bytes)

We can keep reading:

30 82 04 C8 A0 03 02 01 02

Another Sequence embedded in the first one, the TBSCertificate SEQUENCE

TBSCertificate  ::=  SEQUENCE  {
    version         [0]  EXPLICIT Version DEFAULT v1,

The first value should be the version of the certificate:

A0 03

Now this is a different kind of TAG, there are 4 classes of TAGs in ASN.1: UNIVERSAL, APPICATION, PRIVATE, and context-specific. Most of what we use are UNIVERSAL tags, they can be understood by any application that knows ASN.1. The A0 is the [0] (and the following 03 is the length). [0] is a context specific TAG and is used as an index when you have a series of object. The github certificate is a good example of this, because you can see that the next index used is [3] the extensions object:

TBSCertificate  ::=  SEQUENCE  {
    version         [0]  EXPLICIT Version DEFAULT v1,
    serialNumber         CertificateSerialNumber,
    signature            AlgorithmIdentifier,
    issuer               Name,
    validity             Validity,
    subject              Name,
    subjectPublicKeyInfo SubjectPublicKeyInfo,
    issuerUniqueID  [1]  IMPLICIT UniqueIdentifier OPTIONAL,
                         -- If present, version MUST be v2 or v3
                          subjectUniqueID [2]  IMPLICIT UniqueIdentifier OPTIONAL,
                      -- If present, version MUST be v2 or v3
     extensions      [3]  EXPLICIT Extensions OPTIONAL
                      -- If present, version MUST be v3
}

Since those obects are all optionals, skipping some without properly indexing them would have caused trouble parsing the certificate.

Following next is:

02 01 02

Here's how it reads:

  _______ tag:      integer
 |   ____ length: 1 byte
 |  |   _ value:  2
 |  |  |
 |  |  |
 v  v  v
02 01 02 

The rest is pretty straight forward except for IOD: Object Identifier.

Object Identifiers

They are basically strings of integers that reads from left to right like a tree.

So in our Github's cert example, we can see the first IOD is 1.2.840.113549.1.1.11 and it is supposed to represent the signature algorithm.

So go to http://www.alvestrand.no/objectid/top.html and click on 1, and then 1.2, and then 1.2.840, etc... until you get down to the latest branch of our tree and you will end up on sha256WithRSAEncryption.

Here's a more detailed explanation on IOD and here's the microsoft doc on how to encode IOD in DER.

comments (5)

Cryptography Services to audit Let's Encrypt April 2015

Like the audit of OpenSSL wasn't awesome enough, today we learned that we were going to audit Let's Encrypt this summer as well. Pretty exciting agenda for an internship!

https://letsencrypt.org/2015/04/14/ncc-group-audit.html

ISRG has engaged the NCC Group Crypto Services team to perform a security review of Let’s Encrypt’s certificate authority software, boulder, and the ACME protocol. NCC Group’s team was selected due to their strong reputation for cryptography expertise, which brought together Matasano Security, iSEC Partners, and Intrepidus Group.

comment on this story

ASN.1 vs DER vs PEM vs x509 vs PKCS#7 vs .... April 2015

I was really confused about all those acronyms when I started digging into OpenSSL and RFCs. So here's a no bullshit quick intro to them.

PKCS#7

Or Public-Key Crypto Standard number 7. It's just a guideline, set of rules, on how to send messages, sign messages, etc... There are a bunch of PKCS that tells you exactly how to do stuff using crypto. PKCS#7 is the one who tells you how to sign and encrypt messages using certificates. If you ever see "pkcs#7 padding", it just refers to the padding explained in pkcs#7.

X509

In a lot of things in the world (I'm being very vague), we use certificates. For example each person can have a certificate, and each person's certificate can be signed by the government certificate. So if you want to verify that this person is really the person he pretends to be, you can check his certificate and check if the government signature on his certificate is valid.

TLS use x509 certificates to authenticate servers. If you go on https://www.facebook.com, you will first check their certificate, see who signed it, checked the signer's certificate, and on and on until you end up with a certificate you can trust. And then! And only then, you will encrypt your session.

So x509 certificates are just objects with the name of the server, the name of who signed his certificate, the signature, etc...

Example from wikipedia:

    Certificate
        Version
        Serial Number
        Algorithm ID
        Issuer
        Validity
            Not Before
            Not After
        Subject
        Subject Public Key Info
            Public Key Algorithm
            Subject Public Key
        Issuer Unique Identifier (optional)
        Subject Unique Identifier (optional)
        Extensions (optional)
            ...
    Certificate Signature Algorithm
    Certificate Signature

ASN.1

So, how should we write our certificate in a computer format? There are a billion ways of formating a document and if we don't agree on one then we will never be able to ask a computer to parse a x509 certificate.

That's what ASN.1 is for, it tells you exactly how you should write your object/certificate

DER

ASN.1 defines the abstract syntax of information but does not restrict the way the information is encoded. Various ASN.1 encoding rules provide the transfer syntax (a concrete representation) of the data values whose abstract syntax is described in ASN.1.

Now to encode our ASN.1 object we can use a bunch of different encodings specified in ASN.1, the most common one being used in TLS is DER

DER is a TLV kind of encoding, meaning you first write the Tag (for example, "serial number"), and then the Length of the following value, and then the Value (in our example, the serial number).

DER is also more than that:

DER is intended for situations when a unique encoding is needed, such as in cryptography, and ensures that a data structure that needs to be digitally signed produces a unique serialized representation.

So there is only one way to write a DER document, you can't re-order the elements.

And a made up example for an ASN.1 object:

OPERATION ::= CLASS
{
&operationCode INTEGER UNIQUE,

&InvocationParsType,

&ResponseParsAndResultType,

&ExceptionList ERROR OPTIONAL
}

And its DER encoding:

0110 0111 0010 110...

Base64

Base64 is just a way of writing binary data in a string, so you can pass it to someone on facebook messenger for exemple

From the openssl Wiki:

ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/ 
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
0000000000111111111122222222223333333333444444444455555555556666
0123456789012345678901234567890123456789012345678901234567890123

And if you see any equal sign =, it's for padding.

So if the first 6 bits of your file is '01' in base 10, then you will write that as B in plaintext. See an example if you still have no idea about what I'm talking about.

PEM

A pem file is just two comments (that are very important) and the data in base64 in the middle. For example the pem file of an encrypted private key:

-----BEGIN ENCRYPTED PRIVATE KEY-----
MIIFDjBABgkqhkiG9w0BBQ0wMzAbBgkqhkiG9w0BBQwwDgQIS2qgprFqPxECAggA
MBQGCCqGSIb3DQMHBAgD1kGN4ZslJgSCBMi1xk9jhlPxP3FyaMIUq8QmckXCs3Sa
9g73NQbtqZwI+9X5OhpSg/2ALxlCCjbqvzgSu8gfFZ4yo+Xd8VucZDmDSpzZGDod
X0R+meOaudPTBxoSgCCM51poFgaqt4l6VlTN4FRpj+c/WZeoMM/BVXO+nayuIMyH
blK948UAda/bWVmZjXfY4Tztah0CuqlAldOQBzu8TwE7WDwo5S7lo5u0EXEoqCCq
H0ga/iLNvWYexG7FHLRiq5hTj0g9mUPEbeTXuPtOkTEb/0ckVE2iZH9l7g5edmUZ
GEs=
-----END ENCRYPTED PRIVATE KEY-----

And yes the number of - are important

comments (8)

Video: How the RSA attacks using lattices work April 2015

This is my second video, after the first explaining how DPA works. I'm still trying to figure out how to do that but I think it is already better than the first one. Here I explain how Coppersmith used LLL, an algorithm to reduce lattices basis, to attack RSA. I also explain how his attack was simplified by Howgrave-Graham, and the following Boneh and Durfee attack simplified by Herrmann and May as well.

The repo is here, you can check the survey here as well.

Also, follow me on twitter

comment on this story

encryption with a one letter XOR? Really? April 2015

So there is this app that encrypts your data on your mobile, in case it ends up in the wrong hands. Sounds good. And then there is this guy who took a look at it and figured out the data was just XORed with a 128bit keys consisting of only 4s. If the data is longer than 128bits? Let's not encrypt it!

I don't know how legit it is, especially considering how easy it is to just write aes(something) but here you go

comment on this story

Unix command of the day: Tee April 2015

The tee command allow you to write to a file and still display the result in output.

For exemple

ls

display the content of the current folder in stdout (the terminal)

ls > file.txt

saves that in a file file.txt

ls | tee file.txt

saves that in a file file.txt and displays in stdout at the same time

comment on this story

Truecrypt report April 2015

Some news about the Truecrypt open audit: the report is out.

The TL;DR is that based on this audit, Truecrypt appears to be a relatively well-designed piece of crypto software. The NCC audit found no evidence of deliberate backdoors, or any severe design flaws that will make the software insecure in most instances.

comment on this story